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Simplifying (an equation with a fraction within a fraction)

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    I find it difficult to simplify an equation which has a fraction within a fraction. For example, I do not understand how to simplify this:

    \frac{1}{\frac{1}{x+1}+1}

    Apparently, this can be rewritten as this:

    \frac{x+1}{x+2}

    But I do not understand the process used to achieve this. I would be grateful if someone could take me through the process of simplifying this equation and perhaps help me to understand a general method for simplifying equations involving a fraction within a fraction.
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    (Original post by Stikki)
    I find it difficult to simplify an equation which has a fraction within a fraction. For example, I do not understand how to simplify this:

    \frac{1}{\frac{1}{x+1}+1}

    Apparently, this can be rewritten as this:

    \frac{x+1}{x+2}

    But I do not understand the process used to achieve this. I would be grateful if someone could take me through the process of simplifying this equation and perhaps help me to understand a general method for simplifying equations involving a fraction within a fraction.
    If you multiply the entire fraction by (x+1)/(x+1) [which does not change the value since it is equal to 1] you find that the numerator becomes x+1 and the denominator becomes 1+x+1=x+2

    Hence we can write it as \frac{x+1}{x+2}
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    Another way is

    Find a common denominator for the bottom

    \frac{1}{\frac{1+x+1}{x+1}}

    Simplify

    \frac{1}{\frac{x+2}{x+1}}

    Then just remembering division of fractions, 1 divided by a fraction is the same as 1 times the reciprocal.
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    take the denominator:

    (1/(x+1)) + 1

    you can multiply the 1 by (x+1)/(x+1) to make it:

    (1/(x+1)) + (x+1)/(x+1)

    now both fractions in the denominator have a common denominator of x+1, so turn it into one fraction:

    (1/(x+1)) + (x+1)/(x+1) = (1+(x+1))/(x+1)

    which simplifies to (x+2)/(x+1)

    so the fraction is now 1/((x+2)/(x+1))

    as i assume you know, you can take the denominator of the denominator and stick it on the numerator to turn this into:

    (x+1)/(x+2)

    voila
  5. Offline

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    do the smallest one frst, if that makes sense:

    \frac{1}{\frac{1}{x+1}+1}\newlin  e

\implies\frac{1}{\frac{1+x+1}{x+  1}}\newline

\implies\frac{1}{\frac{x+2}{x+1}  }\newline

\implies\frac{x+1}{x+2}
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    I understand how it works now. Thank you everyone.
  7. Offline

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    (Original post by Prokaryotic_crap)
    do the smallest one frst, if that makes sense:

    \frac{1}{\frac{1}{x+1}+1}\newlin  e

\implies\frac{1}{\frac{1+x+1}{x+  1}}\newline

\implies\frac{1}{\frac{x+2}{x+1}  }\newline

\implies\frac{x+1}{x+2}
    Wouldn't it be better to times through by x+1 to begin as it can get confusing with remembering which line is more prevalent?
  8. Offline

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    (Original post by Stikki)
    \frac{1}{\frac{1}{x+1}+1}
    \frac{1}{x+1}+1 = \frac{1}{x+1}+\frac{x+1}{x+1} = \frac{x+2}{x+1}

    Hence, \frac{1}{\frac{1}{x+1}+1}=\frac{  1}{\frac{x+2}{x+1}}

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